What is the integral from-1 to 0 of 6/(1+(2x+1)^2) ?

The integral from-1 to 0 of 6/(1+(2x+1)^2) is (3pi)/2

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integral from-1 to 0 of 6/(1+(2x+1)^2)

\int_{- 1}^{0} \frac{6}{1 + ( 2 x + 1 ) ^{2}} d x

\frac{3 π}{2}

Decimal Notation

4.71238 \dots

Solution steps

\int_{- 1}^{0} \frac{6}{1 + ( 2 x + 1 ) ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{- 1}^{0} \frac{1}{1 + ( 2 x + 1 ) ^{2}} d x

Apply u-substitution

: \int_{- 1}^{1} \frac{1}{2 ( 1 + u ^{2} )} d u

= 6 \cdot \int_{- 1}^{1} \frac{1}{2 ( 1 + u ^{2} )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \frac{1}{2} \cdot \int_{- 1}^{1} \frac{1}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= 6 \cdot \frac{1}{2} [arctan (u)]_{- 1}^{1}

Simplify

6 \cdot \frac{1}{2} [arctan (u)]_{- 1}^{1} : 3 [arctan (u)]_{- 1}^{1}

= 3 [arctan (u)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= 3 \cdot \frac{π}{2}

Simplify

= \frac{3 π}{2}

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What is the integral from-1 to 0 of 6/(1+(2x+1)^2) ?
The integral from-1 to 0 of 6/(1+(2x+1)^2) is (3pi)/2